Question

Show that as the distance from a point source doubles, the sound intensity level decreases by 6 dB.

Answer 1

plez send the attachment also..

incomplete Q

Answer 2

Answer:

The lower in sound depth stage is $6$ $dB$ is proved.

Explanation:

• As we flow similarly alongside the supply, the power emitted with the aid of using the sound supply spreads over a massive and massive area.

• Decibel is used to indicate the sound depth stage.
• $W/m^{2}$ is used to indicate the sound depth.

Inverse-rectangular regulation is used to specify the sound depth. This is given as:

$I$ ∝ $\frac{1}{d^{2} }$

where $d$ is the space from the supply

Step 1:

When the space is doubled, the depth of sound becomes  $\frac{1}{4} th$ of the authentic depth. This is because of inverse rectangular regulation.

So, $\frac{If}{Ii} =\frac{1}{4}$.

Step 2:

As sound depth stage  $=10log(\frac{I}{Io} )dB$

The alternate in sound depth stage while distance doubles become$\beta 2-\beta 1=10log(\frac{1}{4})$

$\beta 2-\beta 1=10* 0.6 dB$

$\beta 2-\beta 1=6$ $dB$

So, the lower in sound depth stage is with the aid of using $6$ $dB$.